First, let's rewrite #ln(x^4)# as #4ln(x),# as #ln(x^a)=aln(x)#
#intx^3ln(x^4)dx=int4x^3ln(x)dx=4intx^3ln(x)dx# (We can factor out #4,# it's just a constant.)
Make the following choices for #u# and #dv:#
#u=ln(x)#
#dv=x^3dx#
Thus, #du# and #v# become:
#du=1/xdx#
#dv=intx^3dx=1/4x^4# (We're not going to include the constant just yet. It can wait until the end.)
Now it becomes apparent why we made the choices we did for #u# and #v.# If we chose #dv=ln(x),# we would have to integrate #ln(x)# for #v,# and would run into our original problem.
Plug in relevant values:
#uv-intvdu=1/4x^4ln(x)-int1/4x^4/xdx#
We can simplify the integral here:
#1/4x^4ln(x)-int1/4x^4/xdx=1/4x^4ln(x)-1/4intx^(cancel(4)3)/cancelx#
Thus, we have
#1/4x^4ln(x)-1/4intx^3dx=1/4x^4ln(x)-(1/4)(1/4)x^4+C=1/4x^4ln(x)-1/16x^4+C#
Let's not forget the #4# we originally factored out...
#4(1/4x^4ln(x)-1/16x^4+C)=x^4ln(x)-1/4x^4+C#
#C# remains unchanged; a constant multiplied by a constant remains a constant.
So,
#intx^3ln(x^4)dx=x^4ln(x)-1/4x^4+C#